Common Fixed Point Theorem in Compact D Metric Spaces

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1 International Mathematical orum, Vol. 6, 2011, no. 13, Common ixed Point Theorem in Compact Metric Spaces 1 Nguyen Van Luong and 2 Nguyen Xuan Thuan epartment of Natural Sciences Hong uc University, Thanh Hoa, Viet Nam 1 luongk6ahd04@yahoo.com, 2 thuannx7@gmail.com Abstract In this paper, we prove a common fixed point theorem for two pairs of weakly compatible mappings in compact metric spaces which satisfy an implicit relation. Mathematics Subject Classification: 47H10, 54H25 Keywords: metric space, compact metric space, implicit relation, weakly compatile 1 Introduction and Preliminaries In 1992, hage [1] introduced the notion of generalized metric or metric spaces and claimed that metric convergence defines a Hausdorff topology and that metric is sequentially continuous in all variables. After that, many author have used these claims to prove fixed point theorems in metric spaces. Unfortunately, almost theorems in metric spaces are not valid (see [4-5]). Recently, Shaban Sedghi et.al [7-10] modify the metric space and defined metric spaces and proved some basic properties and some fixed point and common fixed point theorems in complete metric spaces. In this paper, using the concept of metric space, we prove a common fixed point theorem for two pairs of weakly compatible mappings in compact metric spaces satisfying the implicit relation which be introduced in [2]. Now, we recall some concepts and properties of metric space. efinition 1.1. ([8]) Let X be a nonempty set. A metric on X is a function : X 3 [0, ) that satisfies the following conditions for each x, y, z, a X: (1) (x, y, z) 0, (x, y, z) =0if and only if x = y = z,

2 606 Nguyen Van Luong and Nguyen Xuan Thuan (2) (x, y, z) = (p{x, y, z}), where p is a permutation function, (3) (x, y, z) (x, y, a)+ (a, zz). The pair (X, ) is called a metric space. Example 1.2. ([8]) Let (X, d) be a metric space, then (X, ), with function : X 3 [0, ) be defined by (a) (x, y, z) =d(x, y)+d(y, z)+d(z, x), or (b) (x, y, z) = max{d(x, y),d(y, z),d(z, x)}, for all x, y, z X, is metric space. Remark 1.3. ([8]) In a metric space (X, ), (x, x, y) = (x, y, y) for all x, y X. efinition 1.4. ([8]) Let (X, ) be a metric space. (1) A sequence {x n } in X converges to x if and only if (x n,x n,x)= (x, x, x n ) 0 as n. (2) A sequence {x n } in X is called a Cauchy sequence if for each ε>0, there exists n 0 N such that (x n,x n,x m ) <εfor each n, m n 0. The metric space (X, ) is said to be complete if every Cauchy sequence is convergent. efinition 1.5. A metric (X, ) is said to be (sequentially) compact if every sequence in X has a subsequence that converges to a point in X. Proposition 1.6. Let (X, d) be a compact metric space, then (X, ), with function : X 3 [0, ) be defined by (x, y, z) =d(x, y)+d(y, z)+d(z, x), is a compact metric space. Proof. Let {x n } is a sequence in X. Since (X, d) is compact, there exists subsequence {x nk } of {x n } such that x nk converges to x in (X, d), that is, d(x nk,x) 0ask. We have (x nk,x nk,x)=d (x nk,x nk )+d (x nk,x)+d (x, x nk )=2d (x nk,x) 0, as k. That is, {x nk } converges to x in (X, ). Therefore (X, )isa compact metric space. efinition 1.7. Let (X, 1) and (Y, 2) are two metric spaces. The maping T : X Y is called (sequentially) continuous if {x n } is a sequence in X that converges to x in X, then the sequence {Tx n } converges to Tx in Y. Lemma 1.8. ([8]) Let (X, ) be a metric space. Then is a continuous function in X 3. efinition 1.9. ([3]) Let T and S are two mappings on a set X into itself. Then the mappings are said to be weak compatible if they commute at their coincidence point, that is, Tx = Sx implies that TSx = STx.

3 Common fixed point in - metric spaces Implicit relation Implicit relation on a metric space has been used in many articles (see [2], [6] and references therein). In [6], V. Popa proved a common fixed point theorem in compact metric space for compatible and weakly compatible mappings by using an implicit relation. Very recently, in [2], M. Imdad and J. Ali found out a gap in the proof of the main theorem in [6] and added one essential condition to the implicit relation in [6] and proved a common fixed point theorem for two pair of weakly compatible mappings in compact metric spaces. The following implicit relation be introduced in [2]. Let be the set of functions (t 1,t 2,..., t 6 ):(R + ) 6 R satisfying the following conditions: (1) is non-increasing in variables t 5 and t 6, (2) or every u 0,v >0 (2a) (u, v, v, u, u + v, 0) < 0, or (2b) (u, v, u, v, 0,u+ v) < 0 we have u<v. (3) (u, 0, 0,u,u,0) 0, u >0. (4) (u, u, 0, 0,u,u) 0, u >0. Here is some examples of such functions. Example 2.1. ([2]) efine (t 1,t 2,..., t 6 ):(R + ) 6 R as { (t 1,..., t 6 )=t 1 max t 2,t 3,t 4, t } 5 + t 6. 2 Example 2.2. ([2]) efine (t 1,t 2,..., t 6 ):(R + ) 6 R as { (t 1,..., t 6 )=t 1 max t 2,t 3,t 4, t 5 + t 6,b } t 5 t 6, with b (0, 1). 2 Example 2.3. ([2]) efine (t 1,t 2,..., t 6 ):(R + ) 6 R as (t 1,..., t 6 )=t 3 1 a t 2 3 t2 4 + t2 5 t2 6 1+t 2 + t 3 + t 4, where a (0, 1). Example 2.4. ([2]) efine (t 1,t 2,..., t 6 ):(R + ) 6 R as { (t 1,..., t 6 )=t 1 max t 2, t 3 + t 4, t 5 + t Example 2.5. efine (t 1,t 2,..., t 6 ):(R + ) 6 R as (t 1,..., t 6 )=t 1 t 2. }.

4 608 Nguyen Van Luong and Nguyen Xuan Thuan 3 The main results Our main result, for a compact metric space, reads follows: Theorem 3.1. Let (X, ) a compact metric spaces and A, B, H, G : X X satisfy the following conditions: (i) A(X) G(X) and B(X) H(X), (ii) (A, H) and (B,G) are weakly compatible, (iii) A and H are continuous mappings, (iv) There exists such that (Ax, Ay, Bz), (Hx,Hy,Gz), (Ax, Ay, Hx), (Gz, Bz, Bz), (Hx,Hy,Bz), < 0 (3.1) (Gz, Ay, Ax) for all x, y, x X for which one of (Hx,Hy,Gz), (Ax, Ay, Hx), (Gz, Bz, Bz) is positive. Then A, B, G and H have an unique common fixed point. Proof. We consider the function f : X R which be defined by f(x) = (Ax, Ax, Hx), for all x X Let m = inf x X f(x). There exists a sequence {x n} in X such that m = lim n f(x n ) = lim n (Ax n,ax n,hx n ). Since X is compact, there exists a subsequence {x nk } of {x n } such that lim x n k = x 0 X k Hence, by the continuity of A, H and, we have m = lim (Ax nk,ax nk,hx nk ) n = ( lim Ax nk, lim Ax nk, lim Hx nk, ) k k k = (Ax 0,Ax 0,Hx 0 ) Since A(X) G(X), there exists y 0 X such that Ax 0 = Gy 0. Hence m = (Gy 0,Gy 0,Hx 0 ). We shall show that m = 0. Assume that m>0. Taking x = y = x 0 and z = y in (3.1), we have (Ax 0,Ax 0,By 0 ), (Hx 0,Hx 0,Gy 0 ), (Ax 0,Ax 0,Hx 0 ), (Gy 0,By 0,By 0 ), (Hx 0,Hx 0,By 0 ), < 0 (Gy 0,Ax 0,Ax 0 )

5 Common fixed point in - metric spaces 609 Since (Hx 0,Hx 0,By 0 ) (Hx 0,Hx 0,Gy 0 )+ (Gy 0,By 0,By 0 ) = (Hx 0,Hx 0,Gy 0 )+ (Gy 0,Gy 0,By 0 ) By condition (1), we have (Gy 0,Gy 0,By 0 ), (Gy 0,Gy 0,Hx 0 ), (Ax 0,Ax 0,Hx 0 ), (Gy 0,Gy 0,By 0 ), (Hx 0,Hx 0,Gy 0 )+ < 0 (Gy 0,By 0,By 0 ), 0 or ( (Gy 0,Gy 0,By 0 ),m,m, (Gy 0,Gy 0,By 0 ),m+ (Gy 0,Gy 0,By 0 ), 0) < 0 By the the condition ( 2a), we get (Gy 0,Gy 0,By 0 ) <m (3.2) Since B(X) H(X), there exists z 0 X such that Hz 0 = By 0. rom (3.2), we get (Gy 0,Gy 0,Hz 0 ) <m. Since (Az 0,Az 0,Hz 0 ) m>0, taking x = y = z 0,z = y 0 in (3.1), we have (Az 0,Az 0,By 0 ), (Hz 0,Hz 0,Gy 0 ), (Az 0,Az 0,Hz 0 ), (Gy 0,By 0,By 0 ), (Hz 0,Hz 0,By 0 ), < 0 (Gy 0,Az 0,Az 0 ) Since (Gy 0,Az 0,Az 0 ) = (Az 0,Az 0,Gy 0 ) (Az 0,Az 0,Hz 0 )+ (Hz 0,Gy 0,Gy 0 ) = (Az 0,Az 0,Hz 0 )+ (By 0,By 0,Gy 0 ) by the condition ( 1), we have (Az 0,Az 0,Hz 0 ), (By 0,By 0,Gy 0 ), (Az 0,Az 0,Hz 0 ), (Gy 0,By 0,By 0 ), 0, (Az 0,Az 0,Hz 0 )+ < 0 (By 0,Byy 0,Gy 0 ) By the condition ( 2b), we get (Az 0,Az 0,Hz 0 ) < (By 0,Gy 0,Gy 0 ) (3.3) rom (3.2) and (3.3), we have m (Az 0,Az 0,Hz 0 ) < (By 0,Gy 0,Gy 0 ) <m

6 610 Nguyen Van Luong and Nguyen Xuan Thuan which is a contradiction. Hence m = 0. Therefore That is, (Ax 0,Ax 0,Hx 0 )= (Gy 0,Gy 0,Hx 0 )=0 Ax 0 = Hx 0 = Gy 0 (3.4) Suppose that (Gy 0,Gy 0,By 0 ) > 0, taking x = y = x 0,z = y 0 in (3.1), we have ( (Gy 0,Gy 0,By 0 ), 0, 0, (Gy 0,Gy 0,By 0 ), (Gy 0,Gy 0,By 0 ), 0) < 0 which contradicts with ( 3). Hence (Gy 0,Gy 0,By 0 ) = 0, or rom (3.4) and (3.5), we have Gy 0 = By 0 (3.5) Ax 0 = Hx 0 = Gy 0 = By 0 (3.6) Since (A, H) is weakly compatible, then AHx 0 = HAx 0. Hence A 2 x 0 = AHx 0 = HAx 0 = H 2 x 0. Assume that H 2 x 0 Hx 0, taking x = y = Hx 0,z = y 0 in (3.1), we have (H 2 x 0,H 2 x 0,Gy 0 ), (H 2 x 0,H 2 x 0,Gy 0 ), (H 2 x 0,H 2 x 0,H 2 x 0 ), (Gy 0,By 0,By 0 ), (H 2 x 0,H 2 x 0,By 0 ), (Gy 0,H 2 x 0,H 2 < 0 x 0 ) or ( (H 2 x 0,H 2 x 0,Hx 0 ), (H 2 x 0,H 2 x 0,Hx 0 ), 0, 0, (H 2 x 0,H 2 x 0,Hx 0 ), (H 2 x 0,H 2 x 0,Hx 0 ) ) < 0 which contradicts with (4). Hence H 2 x 0 = Hx 0. Similarly, we can show that G 2 y 0 = Gy 0. Let u = Hx 0 = Gy 0. We have Au = AHx 0 = H 2 x 0 = Hu and Hu = H 2 x 0 = Hx 0 = u. HenceAu = Hu = u. Similarly, we have Gu = Bu = u. Therefore, Au = Hu = Bu = Gu = u, that is, u is a common fixed point of A, B, G and H. Now, we show that u is unique common fixed point. Assume that there exists v X which is also common fixed point of A, B, G, H and v u. Since (Gv, Bv, Bv) > 0, taking x = y = u, z = v in (3.1), we have (Au, Au, Bv), (Hu, Hu, Gv), (Au, Au, Hu), (Gv, Bv, Bv), (Hu,Hu,Bv), < 0 (Gv, Au, Au)

7 Common fixed point in - metric spaces 611 or ( (u, u, v), (u, u, v), 0, 0, (u, u, v), (u, u, v)) < 0 which contradicts with (4). Therefore u = v. Consequently, A, B, G and H have an unique common fixed point. Example 3.2. Let X =[0, 1] with metric (x, y, z) = x y + y z + z x for all x, y, z X. Then (X, ) be a compact - metric space. Let A, B, G, H : X X be given by Ax = x 4,Bx= x 6,Hx= x 2 and Gx = x, for all x X 3 Clearly, A(X) G(X) and B(X) H(X), (A, H) and (B,G) are weakly compatible, A and H are continuous mappings. Let :(R + ) 6 R be given by (t 1,t 2,..., t 6 )=t 1 t 2, for all (t 1,t 2,..., t 6 ) (R + ) 6 then. or all x, y, z X with (Hx,Hy,Gz) > 0, we have (Hx,Hy,Gz) = x 2 y + y 2 2 z + z 3 3 x ( 2 x = 2 4 y + y 4 4 z + z 6 6 x ) 4 = 2 (Ax, Ay, Bz) > (Ax, Ay, Bz) Therefore all conditions of Theorem 3.1 hold and 0 is the unique common fixed point of A, B, G, H. References [1] B.C.hage, Generalised metric spaces and mappings with fixed point, Bull. Calcutta Math. Soc.84(1992),no.4, [2] M. Imdad and J. Ali, Remarks on a common ixed Point Theorem in compact metric spaces, Math. Commun., 15, No. 1, pp (2010). [3] G. Jungck, Common fixed points for noncontinuous nonself maps on nonmetric spaces, ar East J. Math. Sci. 4(1996),

8 612 Nguyen Van Luong and Nguyen Xuan Thuan [4] Naidu S.V.R, Rao K.P.R and Srinivasa Rao N. On the topology of metric spaces and the generation of metric spaces from metric spaces, Internat.J.Math. Math.Sci. 2004(2004), No.51, [5] Naidu S.V.R, Rao K.P.R and Srinivasa Rao N. On the concepts of balls in a metric space, Internat.J.Math.Math.Sci.,2005,No.1(2005) [6] V.Popa, A general fixed point theorem for weakly compatible mappings in compact metric spaces, Turkish J. Math. 25(2001), [7] Shaban Sedghi, K.P.R.Rao and Nabi Shobe, Common fixed point theorems for six weakly compatible mappings in metric spaces, International Journal of Mathematical Sciences. Vol. 6, No. 2 (2007), pp [8] Shaban Sedghi, Nabi Shobe, and Haiyun Zhou, A common fixed point theorem in metric spaces, ixed point Theory and Applications. Volume 2007, Article I 27906, 13 pages. [9] S. Sedghi,. Turkoglu, N. Shobe and S. Sedghi, Common ixed Point Theorems for Six Weakly Compatible Mappings in Metric Spaces, Thai Journal of Mathematics, 7 (2009) Number 2 : [10] S. Sedghi, N. Shobe and S. Sedghi, Common fixed point theorems for two mappings in metric spaces, Journal of Prime Research in Mathematics Vol. 4(2008), Received: September, 2010